Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/LJB01SLASHjones6.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))
F₂(a, empty) -> G₂(a, empty)
F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))
F₂(a, empty) -> G₂(a, empty)
F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

G₂(cons₂(x, k), d) -> G₂(k, cons₂(x, d))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₂(x1, x2) = G₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

G₁ > cons₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

F₂(a, cons₂(x, k)) -> F₂(cons₂(x, a), k)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₂(x1, x2) = F₁(x2)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

F₁ > cons₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(a, empty) -> g₂(a, empty)
f₂(a, cons₂(x, k)) -> f₂(cons₂(x, a), k)
g₂(empty, d) -> d
g₂(cons₂(x, k), d) -> g₂(k, cons₂(x, d))

The set Q consists of the following terms:

f₂(x0, empty)
f₂(x0, cons₂(x1, x2))
g₂(empty, x0)
g₂(cons₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.